i-iw=lt~ problem 13 -...

ID

A

SOLUTION

A

1tfJ = 30° = - rad

6

R = 0.3m

i-iw=lt~

PROBLEM13.61

A thin circular rod is supported in a vertical plane by a bracket at A.Attached to the bracket and loosely wound around the rod is a spring ofconstant k = 40 N/m and undeformed length equal to the arc of circleAB. A 200-g collar C, not attached to the spring, can slide without frictionalong the rod. Knowing that the collar is released ftom rest whenfJ = 30°, determine (a) the maximum height above point B reached bythe collar, (b) the maximum velocity of the collar.

(a) Maximum height

Above B is reached when the velocity at E is.zero

Tc = 0

Point C

(VC)g = WR(I- cosfJ) = (0.2 kg x 9.81m/s2)(0.3m)(l- cos300)

(VE)e = 0

(VE)g = WH = (0.2 x 9.81)(H) = 1.962H (1)

(spring is unattached)

o + 0.4935 + 0.07886 =0 + 0 + 1.962H

H = 0.292m ~

1-1 IN =It'3

PROBLEM 13.61 CONTINUED

(b) The maximum velocity is at B where the potential energy iszero, vB = vrnax

Tc = 0 Vc = 0.4935 + 0.07886 = 0.5724 J

1 2 1( )

2TB = -mvB = - 0.2 kg vrnax2 2

2TB = 0.1vrnax

VB = 0

0+ 0.5724 =(O.I)v~ax

Vrnax= 2.39 m/s ....

---

- ------

ID

A

SOLUTION

R = 0.3m

W = (0.2kg) x (9.81 m/s2)

= 1.962 N

/fw:tt3

PROBLEM 13.62

A thin circular rod is supported in a vertical plane by a bracket at A.Attached to the bracket and loosely wound around the rod is a spring ofconstant k = 40 N/m and undeformed length equal to the arc of circleAB. A 200-g collar C, not attached to the spring, can slide without frictionalong the rod. Knowing that the collar is released from rest at an angle 0with respect to the vertical, determine (a) the smallest value of 0 forwhich the collar will pass through D and reach point A, (b) the velocityof the collar as it reaches point A.

(a) Smallest angle 0 occurs when the velocity at D is close to

zero

Vc = 0

Tc = 0 TD = 0

Point C

MBC = (0.3m)O = 0.30 m

(VCt =~k( MBC)2

(VCt = 1.802

(VC)g = WR(1 - cosO)

(VC)g = (1.962N)(O.3m)(l- cosO)

Vc = (Vct + (VC)g = 1.802 + 0.5886(1 - cosO)

Point D

(VD)e = 0 (spring is unattached)

(VD)g = W(2R) = (2)(1.962 N)(O.3m) = 1.1772J

0+ 1.802 + 0.5586(1- cosO) = 1.1772 J

By trial

(1.8)02 - (0.5886)cosO = 0.5886

o= 0.7522 rad

HW#-3


(b) Velocity at APoint D

VD = 0 TD = 0 VD = 1.1772J (see Part (a»)

Point A

1 2 1( )

2TA=- mvA = - 0.2kg VA2 2

VA = (VA)g = W(R) = (1.962N)(O.3m) = 0.5886J

O.lv~ + 0.5886 = 0 + 1.1772

VA = 2.43 m/s ~

- - -- -- -----

--- -- -- --

1+w~?::'

PROBLEM 13.63

A 6-lb collar can slide without friction on a vertical rod and is resting inequilibrium on a spring. It is pushed down, compressing the spring 6 in.,and released. Knowing that the spring constant is k = 15 lib/in.,determine (a) the maximum height h reached by the collar abbve itsequilibrium position, (b) the maximum velocity of the collar.

--

SOLUTION

~=o(a) Maximum height when Vz = 0

:. 11 = Tz= 0

Position CD

Xl = 6 lb + 6 in. = 0.4 + 6 = 6.4 in.15 lb/in.

(~)l = !kxf = !(15Ib/in.)( 6.4 in.)z2 2

= 307.2lb.in. = 25.6Ib.ft

Position~ (Vg)2 = mgC~ + h) = 6(0.5 + h)

(Ve)z= 0

11 + VI = Tz+ Vz: (Vg)l+ (~)l = (Vg)z+ (~)z

25.6 = 6(0.5 + h)

h = 3.767 ft h = 45.2 in. .....

(b) Maximum velocity occurs when acceleration is 0, equilibrium

position

1 z 1

(6

)z z

13= -mv3 = - - v3 = 0.093167v32 2 32.2

~ = (Vg)3 + (Ve)3 = 6(6) + ~k(XI - 6)z = 36 + 7.5( 6.4 - 6)z

= 37.2lb.in. = 3.llb.ft

11+ VI = T3+ ~: 25.6 = 0.093167vi + 3.1

vrnax= 15.54 ft/s.....

- -


(b) Force of rod on collar AC

Fz = 0 (no friction)

F = Fxi + Fyj

8 = tan-]~ = 14.04°300

Fe =(kMAC)(cos8i + sin8k)

Fe = (320)(0.10923)(cosI4.04°i + sinI4.04°k)

Fe =33.909i+ 8.4797k (N)

2

LF = (Fx + 33.909~i of:(Fy -4~905~j +8.4797k = ~; j +.mgk

(8.5212 m2/s2 )F =4.905N + (0.5)

y 0.15 mFx + 33.909N = 0

Fx = -33.909 N

Fy = 33.309 N

F = -33.9 N i + 33.3 N j ~

-- --

- -

A

SOLUTION

Iv

---

HW:#-?;>

PROBLEM 13.70

A thin circular rod is supported in a vertical plane by a bracket at A.Attached to the bracket and loosely wound aroundthe rod is a spring ofconstant k = 40 N/m and undeformed lengthequalto the arc of circleAB. A 200-g collar C is unattached to the spring and can slide withoutfriction along the rod. Knowing that the collar is released from rest when() = 30°, determine (a) the velocity of the collar as it passes throughpoint B, (b) the force exerted by the rod on the collar as is passesthrough B.

(a) Ve = 0, Te = 0

1 2TB = -mvB2

TB = .!.(0.2 kg)v~2

TB = O.lv~

arc BC = MBe = R(}

MBe = (0.3m)(300 )(Jr)180°

MBe = 0.15708m

(Vet = ~k(MBe)2 = ~(40 N/m)(0.15708 m)2 = 0.49348J

(Ve)g = WR(I- cos(}) = (0.2 kg)(9.81 m/s2)(0.3m)(l- cos300)

(Ve)g = 0.078857J

Ve = (Ve t + (Ve)g = 0.49348 J + 0.078857 J = 0.57234 J

VB = (VBt + (VB)g = 0 + 0 = 0

Te + Ve = TB+ VB; 0 + 0.57234 = O.lv~

v~ = 5.7234 m2/s2 vB = 2.39 m/s ~

(b) + t"f,F =FR - W = mv~R

FR = 1.962 N + (0.2 kg) (5.7234 m2/s2)(0.3m)

FR = 1.962 N + 3.8156 N = 5.7776 N FR = 5.78N ~

(1)

SOLUTION

Loop 1

~ B

B PROBLEM 13.75

(2)

, y..¥",..~."".";,~:

An 8-oz package is projected upward with a velocity v0 by a spring at A;it moves around a frictionless loop and is deposited at C. For each of thetwo loops shown, determine (a) the smallest velocity v0 for which thepackage will reach C, (b) the corresponding force exerted by the packageon the loop just before the package leaves the loop at C.

(a) The smallest velocity at B will occur when the force exerted by thetube on the package is zero.

P-> tt=- 0

~tma

d

I mv2+,'LF=O+mg=-1Lr

AtA

v~ = rg = 1.5 ft (32.2 ft/s2)

v~ = 48.30

1 2TA = -m Vo

2

VA =0(

80Z = 0.5 lb ~ = 0.5 = 0.01553)32.2

1 2 1( )TB = -mvB = -m 48.30 =24.15 m2 2

VB = mg(7.5 + 1.5) = 9mg = 9(0.5) = 4.5Ib.ft

TA+ VA= TB+ VB: !(0.01553)v~ = 24.15(0.01553)+ 4.52 .

AtB

v~ = 627.82 Vo = 25.056 Vo = 25.1 ft/s ~

AtC

1 2 2Tc = -mvc = 0.007765vc2 V: = 7.5mg = 7.5(0.5) = 3.75

TA + VA = Tc + Vc: 0.007765v~ = 0.007765v~ + 3.75

0.007765(25.056)2 - 3.75 = 0.007765v~

v~ = 144.87

--- ---- - ---

Loop 2

B

(2)

-- -


(b)

r~~''mQ,

N = 0.01553(144.87)1.5

N = 1.49989

{Package in tube} N c = 1.500 lb - .....(a) At B, tube supports the package so,

VB ~ 0

VB = 0, TB = 0 VB = mg(7.5 + 1.5)

= 4.5 lb.ft

!(0.01553)v~ = 4.5 => vA = 24.0732

(b) At C

VA = 24.1 ft/s .....

Tc = 0.007765v~, Vc = 7.5mg = 3.75

TA+ VA= Tc + Vc: 0.007765(24.073)2= 0.007765v~ + 3.75

V~ = 96.573

N~-CtJ =0.5

Nc = 0.01553 (96.573

)= 0.99985

1.5

{Package on tube} Nc = 1.000 lb .....

/-Iv #~

PROBLEM 13.83

Knowing that the velocity of an experimental space probe fired from theearth has a magnitude VA =32.5 Mm/h at point A, determine the velocityof the probe as it passes through point B.

SOLUTION

rA = hA + R = 4.3Mm + 6.37Mm

rA = 10.67Mm

rB = hB + R = 72.7Mm + 6.37Mm

\:),

\ rB = 19.07MmAJB_

AtA, VA = 32.5 Mm/h = 9028m/s

TA = .!.m(9028 m/S)2 =40.752 x 106 m2

rA = 10.67 Mm = 10.67 x 106 m

R = 6370km = 637 x lQ6m

(9.81 m/s2)( 6.37 x 106 mt 6

(6

)m = -37.306 x 10 m

10.67x 10 m

AtB

rB = 19.07 Mm =19.07 X 106m

" 2 -

(9.81 m/s2)( 6.37 X 106m) m- 6

(6

)= -20.874 x 10 m

19.07 x 10 m . .

6' 6 12 . 6TA + VA = TB + VB; 40.752 x 10 m - 37.306 x 10 m = -mvB -,20.874 x 10 m

. 2t- . '.

v~ .= 2[40.752 x 106 - 37.306 x 106 + 20.874 x 106.J

VB = 6.9742 x 103m/s =25.107 Mm/h VB = 25.1 Mm/h ...

SOLUTION

I+'V~~

PROBLEM 13.116

A spacecraft of mass m describes a circular orbit of radius 1}around theearth. (a) Show that the additional energy M which must be impartedtothe spacecraft to transfer it to a circular orbit of larger radius r2 is

M = GMm(r2 - r.)21}r2

where M is the mass of the earth. (b) Further show that if the transferfrom one circular orbit to the other circular orbit is executed by placingthe spacecraft on a transitional semielliptic path AB, the amounts ofenergy M A and M B which must be imparted at A and Bare,respectively; proportional to 1}and r2 :

(a) For a circular orbit of radius r

V2 _ GMr

E = T + V = !mv2 _GMm _ 1 GMm2 r - -2~ (1)

Thus M required to pass from circular orbit of radius 1}to circularorbit of radius r2 is

M = GMm(r2 - 1})21}r2

(2) (Q.E.D.)

(b) For an elliptic orbit we recall Equation (3) derived in

Problem 13.113 (with vp =VI)

B At point A: Initially spacecraft is in a circular orbit of radius 1}

V2. _ GMclrc - -1}

~irc = !mv2. - 1 GM2 clrc - -m-2 YiI

---- - ---

-- ~ -- ---


After the spacecraft engines are fIred and it is placed on asemi-elliptic path AB, we recall

2 2GM r2VI =. .-

('i + r2) rl

T. _ 1 2 _ 1 2GMr2I - - mVI - - m

2 2 rl (rl + r2)And

At point A, the increase in energy is

1 2GMr2 1 GMM A =11 - I::irc = - m

( )- - m-

2 'i 'i + r2 2 rl

Recall Equation (2): (Q.E.D)

A similar derivation at point B yields,

(Q.E.D)

i-iw=lt~ problem 13 -...

Documents